The first two chi square tests were conducted in order to see wheather there is any difference between samples of jelly babies in the same habitat ( first in woodland, than in grassland). The third test is done in order to investigate whether the samples of jelly babies differ significantly between the woodland and grassland habitats.
Chi square test 1
Null hypothesis : The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of woodland.
Observed frequencies in woodland
Black
Green
Orange
Red
Yellow
Pop 1
4
4
1
3
8
Pop2
0
8
1
3
8
Pop3
4
5
3
2
6
Total
8
17
5
8
22
Expected frequencies in woodland
Black
Green
Orange
Red
Yellow
Pop1
8/3
17/3
5/3
8/3
22/3
Pop2
8/3
17/3
5/3
8/3
22/3
Pop3
8/3
17/3
5/3
8/3
22/3
Since the expected frequency of black, orange and red jelly babies is less than three, black and green jelly babies should be combined together in one group and orange, red and yellow jelly babies into another group due to their physiology and predation similarities.
Observed frequencies after combining
Black & Green
Orange, Red & Yellow
Pop1
8
12
Pop2
8
12
Pop3
9
11
Total
25
35
Expected frequencies after combining
Black & Green
Orange, Red & Yellow
Pop1
25/3
35/3
Pop2
25/3
35/3
Pop3
25/3
35/3
Our calculated value (χ²) = 67/525 = 0.12761904 Degrees of freedom (df) = (3-1) x (2-1) = 2 Critical value (P=0.05, 2 df) = 5.99
Conclusion: As our calculated χ² value is smaller than the critical value we will accept our Null hypothesis which states that there is no difference in the proportion of differently coloured jelly babies between the three populations of woodland. We can conclude that we would see deviations from the expected values as large as this in more than 5% occasions if we repeated our experiment.
Chi square test 2
Null hypothesis: The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of grassland.
Observed frequencies in grassland
Black
Green
Orange
Red
Yellow
Pop 1
0
5
4
0
11
Pop2
2
8
0
0
10
Pop3
1
9
1
1
8
Total
3
22
5
1
29
Expected frequencies in grassland
Black
Green
Orange
Red
Yellow
Pop1
1
22/3
5/3
1/3
29/3
Pop2
1
22/3
5/3
1/3
29/3
Pop3
1
22/3
5/3
1/3
29/3
As the expected frequencies of black, orange and red jelly babies are less than three these will be combined with green and yellow jelly babies according to their physiology and predation.
Observed frequencies after combining
Black & Green
Orange, Red & Yellow
Pop1
5
15
Pop2
10
10
Pop3
10
10
Total
25
35
Expected frequencies after combining
Black & Green
Orange, Red & Yellow
Pop1
25/3
35/3
Pop2
25/3
35/3
Pop3
25/3
35/3
Our calculated value ( χ²) = 3.428571 Degrees of freedom (df) = (3-1) x (2-1) = 2 Critical value (P=0.05, 2 df) = 5.99
Conclusion: As our calculated χ² value is smaller than the critical value we will again accept our Null hypothesis.
Chi square test 3
Null Hypothesis: There is no difference in the proportion of black, green, orange, red and yellow jelly babies between the woodland and grassland.
Observed Frequency
Black
Green
Orange
Red
Yellow
Wood
4
4
1
3
8
Wood
0
8
1
3
8
Grass
0
5
4
0
11
Grass
2
8
0
0
10
Grass
1
9
1
1
8
Wood
4
5
3
2
6
Total
11
39
10
9
51
Expected Frequency
Black
Green
Orange
Red
Yellow
Wood
11/6
6.5
10/6
9/6
8.5
Wood
11/6
6.5
10/6
9/6
8.5
Grass
11/6
6.5
10/6
9/6
8.5
Grass
11/6
6.5
10/6
9/6
8.5
Grass
11/6
6.5
10/6
9/6
8.5
Wood
11/6
6.5
10/6
9/6
8.5
As the expected frequencies of black, orange and red jelly babies are all less than 3.0 these will be combined with green and yellow once based on their physiology and predation.
Observed Frequencies after combining
Black & Green
Orange, Red &Yellow
Wood
8
12
Wood
8
12
Grass
5
15
Grass
10
10
Grass
10
10
Wood
9
11
Total
50
70
Expected Frequencies after combining
Black & Green
Orange, Red &Yellow
Wood
50/6
70/6
Wood
50/6
70/6
Grass
50/6
70/6
Grass
50/6
70/6
Grass
50/6
70/6
Wood
50/6
70/6
Our calculated value ( χ²) = 3.56571428
Degrees of freedom (df) = (6-1) x (2-1) = 5
Critical value from the statistical table ( P=0.05 and 1 df ) is 11.07
Conclusion: Since our calculated χ² value is less than the critical value we can accept our Null Hypothesis.
1
Black Green Orange Red Yellow
4 4 1 3 8
2
Green Orange Red Yellow
8 1 3 8
3
Green Orange Yellow
5 4 11
4
Black Green Yellow
2 8 10
5
Black Green Orange Red Yellow
1 9 1 1 8
6
Black Green Orange Red Yellow
4 5 3 2 6
The first two chi square tests were conducted in order to see wheather there is any difference between samples of jelly babies in the same habitat ( first in woodland, than in grassland). The third test is done in order to investigate whether the samples of jelly babies differ significantly between the woodland and grassland habitats.
Chi square test 1
Null hypothesis : The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of woodland.
Our calculated value (χ²) = 67/525 = 0.12761904
Degrees of freedom (df) = (3-1) x (2-1) = 2
Critical value (P=0.05, 2 df) = 5.99
Conclusion: As our calculated χ² value is smaller than the critical value we will accept our Null hypothesis which states that there is no difference in the proportion of differently coloured jelly babies between the three populations of woodland. We can conclude that we would see deviations from the expected values as large as this in more than 5% occasions if we repeated our experiment.
Chi square test 2
Null hypothesis: The proportion of black, green, orange, red and yellow jelly babies does not differ between the three populations of grassland.
Our calculated value ( χ²) = 3.428571
Degrees of freedom (df) = (3-1) x (2-1) = 2
Critical value (P=0.05, 2 df) = 5.99
Conclusion: As our calculated χ² value is smaller than the critical value we will again accept our Null hypothesis.
Chi square test 3
Null Hypothesis: There is no difference in the proportion of black, green, orange, red and yellow jelly babies between the woodland and grassland.
Our calculated value ( χ²) = 3.56571428
Degrees of freedom (df) = (6-1) x (2-1) = 5
Critical value from the statistical table ( P=0.05 and 1 df ) is 11.07
Conclusion: Since our calculated χ² value is less than the critical value we can accept our Null Hypothesis.